math (final trig problem)
posted by jacob .
Solve for x on the interval [0,π/2):
cos^3(2x) + 3cos^2(2x) + 3cos(2x) = 4
I havent done trig for a while so what exactly does that mean in solving for x on that interval and how would i go about doing that?
Where are you getting these tough trig questions??
how about letting cos 2x = t
then your equation becomes
t^3 + 3t^2 + 3t  4 = 0
using Newton's Method I was able to come up with ONE real root , t=.709976
so cos 2x = .709976
for 2x= .7813322 radians
and finally by dividing by 2
x = appr. .39
You can factorize the cubic equation Reiny gave above:
t^3 + 3t^2 + 3t  4 = 0 >
(t+1)^3 = 5 >
t = 1 + 5^(1/3)
Count, now that was clever.
good for you to recognize the sequence
1,3,3,(1) in Pascal's triangle.
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